LTS Termination Proof

by T2Cert

Input

Integer Transition System

Initial Location: 6

Transitions: (pre-variables and post-variables)

0	0	1:	− i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
0	1	1:	− i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
2	2	0:	___const_10_0 − i_0 ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
2	3	3:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − ___const_10_0 + i_0 ≤ 0 ∧ −1 − i_0 + i_post ≤ 0 ∧ 1 + i_0 − i_post ≤ 0 ∧ i_0 − i_post ≤ 0 ∧ − i_0 + i_post ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
3	4	2:	− i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
1	5	4:	− i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
5	6	3:	0 ≤ 0 ∧ 0 ≤ 0 ∧ i_post ≤ 0 ∧ − i_post ≤ 0 ∧ i_0 − i_post ≤ 0 ∧ − i_0 + i_post ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
6	7	5:	− i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

− i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

and for every transition t, a duplicate t is considered.

2 Transition Removal

We remove transitions 0, 1, 2, 5, 6, 7 using the following ranking functions, which are bounded by −17.

6:	0
5:	0
2:	0
3:	0
0:	0
1:	0
4:	0
6:	−7
5:	−8
2:	−9
3:	−9
3_var_snapshot:	−9
3*:	−9
0:	−10
1:	−11
4:	−12

Hints:

9	lexWeak[ [0, 0, 0, 0, 0, 0] ]
3	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0] ]
4	lexWeak[ [0, 0, 0, 0, 0, 0] ]
0	lexStrict[ [0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0] ]
1	lexStrict[ [0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0] ]
2	lexStrict[ [0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0] ]
5	lexStrict[ [0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0] ]
6	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]
7	lexStrict[ [0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0] ]

3 Location Addition

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

3* 11 3: − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

4 Location Addition

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

3 9 3_var_snapshot: − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

5 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

5.1 SCC Subproblem 1/1

Here we consider the SCC { 2, 3, 3_var_snapshot, 3* }.

5.1.1 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 2.

2:	−1 + 4⋅___const_10_0 − 4⋅i_0
3:	1 + 4⋅___const_10_0 − 4⋅i_0
3_var_snapshot:	4⋅___const_10_0 − 4⋅i_0
3*:	2 + 4⋅___const_10_0 − 4⋅i_0

Hints:

9	lexWeak[ [0, 0, 0, 4, 4, 0] ]
11	lexWeak[ [0, 0, 0, 4, 4, 0] ]
3	lexStrict[ [0, 0, 0, 0, 4, 0, 4, 4, 0] , [0, 0, 4, 0, 0, 0, 0, 0, 0] ]
4	lexWeak[ [0, 0, 0, 4, 4, 0] ]

5.1.2 Transition Removal

We remove transitions 9, 4 using the following ranking functions, which are bounded by −2.

2:	−2
3:	0
3_var_snapshot:	−1
3*:	1

Hints:

9	lexStrict[ [0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0] ]
11	lexWeak[ [0, 0, 0, 0, 0, 0] ]
4	lexStrict[ [0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0] ]

5.1.3 Transition Removal

We remove transition 11 using the following ranking functions, which are bounded by −1.

2:	0
3:	−1
3_var_snapshot:	0
3*:	0

Hints:

11	lexStrict[ [0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0] ]

5.1.4 Splitting Cut-Point Transitions

We consider 1 subproblems corresponding to sets of cut-point transitions as follows.

5.1.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 8.

5.1.4.1.1 Splitting Cut-Point Transitions

There remain no cut-point transition to consider. Hence the cooperation termination is trivial.

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